This is one of the question I got for my Calculus midterm. I had it a few days ago and I still don't know the answer so I am hoping some calculus geniuses on this site can help me figure it out.
Q: Let R be the region below the function $f(x) = e^{-2x^2-2x}$ and above the x-axis for $1 \le x \le 4$. Find the volume of the solid obtained by fully revolving R about the line $x=-1$.
I attempted shell method:
$$V = \int_1^2 2\pi (x+1) e^{-2x^2-2x} dx$$
But then I don't know how to integrate this. So I tried using cylinder method by finding the inverse of $f(x)$. After interchanging $x$ and $y$ and simplifying it, I got:
$$f^{-1}(x) = \frac{\sqrt{1-\ln(x^2)}-1}{2}$$ (Tell me if I am wrong.)
Therefore,
$$V = \int_{e^{-24}}^{e^-4} \pi \left[\left(\frac{\sqrt{1-\ln(x^2)}-1}{2} + 1\right)^2 -1\right]dx$$
which I also don't know how to integrate.
Can anyone please show me how to integrate any of these two? Or if I am approaching this problem incorrectly, please feel free to correct me.